Binomial Probability Distribution on TI89
Copyright © 2007–2020 by Stan Brown
Copyright © 2007–2020 by Stan Brown
Summary: With your TI89/92, you can do all types of probability calculations for a binomial probability distribution.
See also: TI83/84 users can use the program in MATH200A part 3 or the calculator procedure here, in Stats without Tears, to compute binomial probability.
To compute the binomial probability for
one particular number of successes, use the binompdf
function.
TI89 Home Screen  TI89 Stats/List Editor  

Keystrokes  [CATALOG ] [F3 ] [plain ( makes B ] [▼ ] 
[F5 ] [ALPHA ( makes B ] 
Format  binompdf(n, p, x)  (dialog box) 
Notice the difference: When you’re in the catalog, the
calculator puts you into alpha mode, so you don’t press
[ALPHA ] to select a letter. But when you’re in the
Stats/List editor, you do press [ALPHA ] to select a
letter.

Example 1: Larry’s batting average is .260. If he’s at bat four times, what is the probability that he gets exactly two hits?
Solution:
n = 4, p = 0.26, x = 2
Note: Some textbooks use r for number of successes, rather than x.
binompdf(4,.26,2)
= 0.2221
To compute the binomial probability for a
range of numbers of successes from xlow to
xhigh, use the
binomcdf
function.
TI89 Home Screen  TI89 Stats/List  

Keystrokes  [CATALOG ] [F3 ] [plain ( makes B ] 
[F5 ] [ALPHA ) makes C ] 
Format  binomcdf(n, p, xlow, xhigh)  (dialog box) 
Example 2: Larry’s batting average is .260. If he’s at bat six times, what is the probability that he gets two to four hits?
Solution:
n = 6, p = 0.26, 2 ≤ x ≤ 4
binomcdf(6,.26,2,4)
= 0.4840
Example 3: Suppose 65% of the registered voters in Dryden are Republicans. In a random sample of ten registered voters, what’s the probability of fewer than six Republicans?
Solution: “Fewer than six” is zero through five.
n = 10, p = 0.65, 0 ≤ x ≤ 5
binomcdf(10, .65, 0, 5) = 0.2485
There’s about one chance in four of getting fewer than six Republicans in a random sample of ten registered voters.
Example 4: With the same data, what’s the probability of getting eight or more Republicans?
Solution: “Eight or more” means eight to ten since there are only ten trials.
binomcdf(10,.65,8,10)
= 0.2616
Example 5: A fair die has a 1/6 chance of rolling a 2. In 24 rolls, what's the probability of getting no more than three 2’s?
Solution: “No more than three” means zero to three.
n = 24, p = 1/6, 0 ≤ x ≤ 3
binomcdf(24,1/6,0,3)
= 0.4155
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